Chemical Balance Exercises

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Chemical balance is one of the subjects that most fall in Enem and entrance exams.

Aspects of reversible reactions are addressed in the questions and the candidates are evaluated both by calculations and by the concepts that involve this theme.

With that in mind, we made this list of questions with different approaches to chemical balance.

Take advantage of the resolution comments to prepare for the exams, and check out the step-by-step instructions on how to solve the questions.

General concepts of chemical equilibrium

1. (Uema) In the equation aA space plus space bB harpoon space right over harpoon left 2 to 1 space cC space plus space dD, after reaching chemical equilibrium, we can conclude the equilibrium constant straight K with straight c subscript space equal to space numerator space left square bracket C right square bracket to the power of straight c space. space left square bracket D right square bracket to the power of straight d over denominator left square bracket straight The right square bracket to the power of straight to space. space left square bracket straight B right square bracket to the power of straight b end of fraction, about which it is correct to state that:

a) the higher the value of Kc, the lower the yield of the direct reaction.
b) Kç regardless of temperature.
c) if the speeds of the forward and inverse reactions are equal, then Kc = 0.
d) Kç it depends on the initial molarities of the reactants.
e) the greater the value of Kc, the greater the concentration of the products.

Correct answer: e) the greater the value of Kc, the greater the concentration of the products.

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The direct reaction is represented by the number 1, where: aA space plus space bB space right arrow with 1 superscript space cC space plus space dD

The inverse reaction is represented by aA space plus space bB space left arrow of 2 space space cC space more space dD

The value of Kç it is calculated by the ratio between the concentrations of products and reagents.

straight K with straight c subscript space equal to space numerator space left square bracket C right square bracket to the power of straight c space. space left square bracket D right square bracket to the power of straight d over denominator left square bracket straight The right square bracket to the power of straight to space. space left square bracket straight B right square bracket to the power of straight b end of fraction

The numerator (which contains the products) is directly proportional to the equilibrium constant. Therefore, the higher the value of Kç, the greater the yield of the direct reaction, as more product is being formed and, consequently, the greater the concentration of products.

The value of Kç varies with temperature, because when we change its value, the endothermic (heat absorption) or exothermic (heat release) reaction can be favored and, with that, more reagent or product can be consumed or created, thus changing the equilibrium constant that depends on the concentration of reagents.

Kc depends on the molar quantities of the components when equilibrium is established and when the rates of the forward and reverse reactions are equal.

2. (UFRN) The chemical balance is characterized by being dynamic at the microscopic level. To obtain quantitative information on the extent of chemical equilibrium, the equilibrium constant quantity is used. Consider the following strip:

chemical balance

Applied to chemical balance, the character's idea of ​​balance:

a) It is correct because, in chemical equilibrium, half the amounts are always products, and the other half are reactants.
b) It is not correct, since, in chemical equilibrium, the concentrations of products and those of reactants can be different, but they are constant.
c) It is correct because, in chemical equilibrium, the concentrations of reactants and products are always the same, as long as the equilibrium is not disturbed by an external effect.
d) It is not correct, since, in chemical equilibrium, the concentrations of the products are always higher than those of the reactants, as long as the equilibrium is not affected by an external factor.
e) It is correct because, in chemical equilibrium, the concentrations of reactants and products are not always the same.

Correct answer: b) It is not correct, since, in chemical equilibrium, the concentrations of products and those of reactants can be different, but they are constant.

At equilibrium, the quantities of products and reagents can be calculated based on the constant of balance, and not necessarily should be half the amount of products and the other half reagents.

Equilibrium concentrations are not always the same, they may be different, but constant if no disturbances occur in equilibrium.

Equilibrium concentrations must depend on which reaction is being favored, whether the forward or the reverse. We can know this by the value of Kç: if Kçbigger then 1, Direct reaction is favored. already if Kç less than 1 the reverse reaction is favored.

Chemical Balance Charts

3. (UFPE) At the beginning of the 20th century, the expectation of the First World War generated a great need for nitrogen compounds. Haber pioneered the production of ammonia from nitrogen in the air. If ammonia is placed in a closed container, it decomposes according to the following unbalanced chemical equation: NH3(g) → N2(g) + H2(g). The variations in concentrations over time are illustrated in the following figure:

chemical balance chart

From the analysis of the figure above, we can state that curves A, B and C represent the temporal variation of the concentrations of the following reaction components, respectively:
a) H2, no2 and NH3
b) NH3, H2 and no2
c) NH3, no2 and H2
d) No2, H2 and NH3
e) H2, NH3 and no2

Correct answer: d) N2, H2 and NH3.

1st step: balance the chemical equation.

2 NH3(g) → N2(g) + 3 H2(g)

With the balanced reaction, we realized that it takes 2 moles of ammonia to decompose into nitrogen and hydrogen. Also, the amount of hydrogen produced in the reaction is three times greater than that of ammonia.

2nd step: interpret chart data.

If ammonia is being decomposed, then in the graph its concentration is maximum and decreases, as seen in curve C.

The products, as they are being formed, at the beginning of the reaction the concentrations are zero and increase as the reactant becomes a product.

Since the amount of hydrogen produced is three times greater than that of nitrogen, then the curve for this gas is the largest, as noted in B.

The other product being formed is nitrogen, as seen in curve A.

4. (Cesgranrio) The system represented by the equation straight F space more space straight G space right arrow on left arrow straight space H was in balance. The equilibrium state was abruptly altered by an addition of substance G. The system reacts to restore balance. Which of the following charts best represents the changes that occurred during the process described?

balance shift graphics

Correct answer: d).

balance disturbance graph

As the system was in equilibrium at the beginning, the amounts of substances G and H remained constant.

The disturbance occurred because the concentration of G increased and the system reacted by transforming this reactant in more product H, shifting the balance to the right, that is, favoring the reaction direct.

We observe that the reagent curve G decreases as it is being consumed, and the product curve H increases as it is being formed.

When a new equilibrium is established, quantities become constant again.

Equilibrium constant: relationship between concentration and pressure

5. (UFRN) Knowing that KP = Kç (RT)n, we can say that KP = Kç, for:

steel2(g) + H2(g) ↔ CO(g) + H2O(g)
b) H2(g) + ½ the2(g) ↔ H2O(1)
c) No2(g) + 3 H2(g) ↔ 2 NH3(g)
d) NO(g) + ½ O2(g) ↔ NO2(g)
e) 4 FeS(s) + 7 O2(g) ↔ 2 Fe2O3(s) + 4 SO2(g)

Correct answer: a) CO2(g) + H2(g) ↔ CO(g) + H2O(g)

To KP be equal to Kç the variation in the number of moles must equal zero, as any number raised to zero results in 1:

KP = Kç (RT)0
KP = Kç x 1
KP = Kç

The change in the number of moles is calculated by:

∆n = Number of moles of products - Number of moles of reagents

In this calculation, only the coefficients of substances in the gaseous state participate.

Applying to each alternative equation, we have:

steel2(g) + H2(g) ↔ CO(g) + H2O(g) ∆n = [(1+1) - (1+1)] = 2 - 2 = 0
b) H2(g) + ½ the2(g) ↔ H2O(1) ∆n = [0 - (1+1/2)] = 0 - 3/2 = - 3/2
c) No2(g) + 3 H2(g) ↔ 2 NH3(g) ∆n = [2 - (1+3)] = 2 - 4 = - 2
d) NO(g) + ½ the2(g) ↔ NO2(g) ∆n = [1 - (1+1/2)] = 1 - 3/2 = - 1/2
e) 4 FeS(s) + 7 O2(g) ↔ 2 Fe2O3(s) + 4 SO2(g) ∆n = [(0+4) - (0+7)] = 4 - 7 = - 3

With these results, we can observe that the alternative whose value corresponds to the required result is the one in the first equation.

6. (UEL-adapted) For the reaction represented by 3 space Fe with left parenthesis s right parenthesis subscript end of subscript space plus space 4 space straight H with 2 subscript straight O with left parenthesis straight g right parenthesis subscript end of subscript harpoon space right over harpoon left space Fe with 3 subscript straight O with 4 left parenthesis straight s right parenthesis subscript end of subscript space plus space 4 straight space H with 2 left parenthesis straight g right parenthesis subscript end of subscript spacethe equilibrium constants Kç and KP are expressed by the equations: (Given: p = partial pressure)

straight to right parenthesis square space K with straight c subscript space equal to numerator left square bracket straight H with 2 subscript right square bracket space. space left square bracket Fe with 3 subscript square O with 4 subscript right square bracket on denominator left square bracket Fe right square bracket. space left square bracket H with 2 straight subscript The right square bracket end of fraction square space and square space K with straight p subscript space equal to p to the power of 4 straight H with 2 subscript straight b right parenthesis space K with straight c subscript space equal to numerator left parenthesis Fe with 3 subscript straight O with 4 subscript straight parenthesis right on denominator left square bracket Fe right square bracket to cube end of fraction straight space and square space K with straight p subscript space equal to p straight space H with 2 subscript straight O straight c right parenthesis space straight K with straight c subscript space equal to numerator left parenthesis straight H with 2 subscript right square bracket to the power of 4 space. space left square bracket Fe with 3 subscript square O with 4 subscript right square bracket on denominator left square bracket Fe right square bracket cubed. space left square bracket straight H with 2 straight subscript The right square bracket to the power of 4 of the fraction square space and square space K with straight p subscript space equal to numerator p italic space Fe over denominator p italic space Fe with 3 straight subscript O with 4 subscript end of fraction straight d right parenthesis space K with straight c subscript space equal to numerator left parenthesis straight H with 2 subscript right parenthesis space. space left square bracket Fe with 3 subscript O with 4 subscript right square bracket on denominator left square bracket H with 2 square subscript O right bracket to the power of 4 end of the fraction straight space and straight space K with straight p subscript space equal to numerator p to the power of 4 straight H with 2 subscript space. p italic space Fe space with 3 straight subscript O with 4 subscript on denominator p to the power of 4 straight H with 2 straight subscript O space. space p to the power of italic 3 italic space Fe end of fraction straight and right parenthesis straight space K with straight c subscript space equal to numerator left square bracket H with 2 subscript right square bracket to the power of 4 on denominator left square bracket H with 2 square subscript Right bracket to the power of 4 end of the fraction space straight and straight space K with straight p subscript space equal to numerator p to the power of 4 straight H with 2 subscript over denominator p to the power of 4 straight H with 2 straight subscript The end of fraction

Correct alternative: straight and right parenthesis square space K with straight c subscript space equal to numerator left square bracket straight H with 2 subscript right square bracket à power of 4 over denominator left square bracket H with 2 straight subscript The right square bracket to the power of 4 end of the fraction square space and space straight K with straight p subscript space equal to numerator p to the power of 4 straight H with 2 subscript over denominator p to the power of 4 straight H with 2 straight subscript The end of the fraction

The equilibrium constant is calculated by: straight K with straight c subscript space equal to space numerator space left square bracket C right square bracket to the power of straight c space. space left square bracket D right square bracket to the power of straight d over denominator left square bracket straight The right square bracket to the power of straight to space. space left square bracket straight B right square bracket to the power of straight b end of fraction

Solid compounds, due to their constant concentrations, do not participate in the calculation of Kç, therefore, the equilibrium constant for the given equation is: straight K with straight c subscript space equal to numerator parenthesis left straight straight H with 2 subscript right parenthesis to power of 4 over denominator left square bracket H with 2 square subscript The right square bracket to the power of 4 end of the fraction space

For the equilibrium constant, in terms of pressure, only the gases participate in the calculation, so: straight K with straight p subscript space equal to numerator p to the power of 4 straight H with 2 subscript over denominator p to the power of 4 straight H with 2 straight subscript The end of the fraction

Calculation of the equilibrium constant

7. (Enem/2015) Several acids are used in industries that dispose of their effluents in water bodies, such as rivers and lakes, which can affect the environmental balance. To neutralize the acidity, calcium carbonate salt can be added to the effluent, in appropriate amounts, as it produces bicarbonate, which neutralizes the water. The equations involved in the process are presented:

equilibrium reactions

Based on the values ​​of the equilibrium constants of reactions II, III, and IV at 25°C, what is the numerical value of the equilibrium constant of reaction I?

a) 4.5 x 10-26
b) 5.0 x 10-5
c) 0.8 x 10-9
d) 0.2 x 105
e) 2.2 x 1026

Correct answer: b) 5.0 x 10-5

1st step: use Hess's law to make the necessary adjustments.

Given a chemical equation: aA space more space bB space right arrow cC space more space dD

The constant is calculated by: straight K space equal to numerator space left square bracket straight C right square bracket to the power of straight c space. space left square bracket D right square bracket to the power of straight d over denominator left square bracket straight The right square bracket to the power of straight to space. space left square bracket straight B right square bracket to the power of straight b end of fraction

But if we reverse the equation, we get as a result: cC space more space dD space right arrow space aA space more space bB

And the constant becomes the inverse: straight K apostrophe space equal to space 1 over straight K

To arrive at equation 1, given in the question, we need to invert equation II, as in the previous example.

2nd step: Manipulate equations II, III and IV to arrive at the result of equation I.

Eq apostrophe left parenthesis II right parenthesis two dots space space space diagonally strike out over straight H to the most end power of strike out space plus space diagonal strike up over CO with 3 subscript to the power of 2 minus end of exponential end of strikeout space right arrow over left arrow space HCO with 3 subscript to the power of minus space inverse space space space Eq space left parenthesis II right parenthesis Eq space left parenthesis III right parenthesis colon space CaCO space with 3 subscript space right arrow on left arrow space Ca at the power of 2 more end of exponential space more space crossed out diagonally up over CO with 3 subscript to the power of 2 minus end of exponential end of strikeout Eq space left parenthesis IV right parenthesis colon space CO with 2 subscript space plus straight space H with 2 straight subscript Space right arrow over left arrow space crossed out diagonally up over straight H to the power of the most end of the strikeout space more HCO space with 3 subscript to the minus power in lower frame closes frame Eq space left parenthesis straight I right parenthesis colon space space space CaCO space with 3 subscript space plus CO space with 2 subscript space plus straight space H with 2 straight subscript Space right arrow over left arrow space 2 HCO with 3 subscript à minus power

3rd step: calculate the equilibrium constant of equation I.

Calculating KI is done by multiplying the constant values.

straight K with straight I subscript space equal to straight space K apostrophe with II subscript straight space x straight space K with III subscript straight space x straight space K with IV straight subscript K with straight I subscript space equal to 1 over straight K with II straight subscript x straight space K with III subscript straight space x straight space K with IV straight subscript K with straight I subscript space equal to numerator 1 over denominator 3 straight space x space 10 to the minus 11 end power of the exponential end of the fraction multiplication sign space 6 straight space x space 10 to the minus power 9 end of exponential straight space x space 2 comma 5 straight space x space 10 to the power minus 7 end of straight exponential K with straight I subscript space equal to numerator 6 straight space x space 10 to minus 9 end of exponential straight space x space 2 comma 5 straight space x space 10 to the minus 7th power of the exponential over denominator 3 straight space x 10th space to the minus 11th power of the exponential end of fraction

As in the calculation we have equal powers of bases, we repeat the base and add the exponents.

straight K with straight I subscript space equal to numerator 15 straight space x space 10 to the power of minus 9 plus left parenthesis minus 7 right parenthesis end of exponential over denominator 3 straight space x space 10 to the power of minus 11 end of exponential end of fraction straight K with straight I subscript space equal to numerator 15 straight space x 10 space to minus 16 end power of exponential over denominator 3 straight space x 10 space to minus 11 end power of exponential end of fraction

Since we now have a division with equal powers of bases, we repeat the base and subtract the exponents.

straight K with straight I subscript space equals space space 5 straight space x space 10 to the power of minus 16 minus left parenthesis minus 11 right parenthesis end of straight exponential K with straight I subscript space equals space space 5 straight space x space 10 to the minus 16 power plus 11 end of straight exponential K with straight I subscript space equal to space 5 space straight space x space 10 to the power of minus 5 end of exponential

8. (UnB) Phosphorus pentachloride is a very important reagent in organic chemistry. It is prepared in the gas phase through the reaction: 1 PCl space with 3 left parenthesis straight g right parenthesis subscript end of subscript space plus 1 Cl space with 2 left parenthesis straight g right parenthesis subscript end of subscript space right arrow over left arrow space 1 PCl space with 5 left parenthesis straight g right parenthesis subscript end of subscribed
A 3.00 L capacity flask contains at equilibrium, at 200 °C: 0.120 mol PCl5(g), 0.600 mol of PCl3(g) and 0.0120 mol of CL2(g). What is the value of the equilibrium constant at this temperature?

Correct answer: 50 (mol/L)-1

1st step: Assemble the expression of the equilibrium constant for the reaction.

straight K with straight c subscript space equal to space numerator left square bracket Products right square bracket on denominator left square bracket Reagents right square bracket end of fraction equal to numerator left square bracket PCl with 5 subscript right bracket on denominator left square bracket PCl with 3 subscript right bracket right space x space left square bracket Cl with 2 subscript right bracket end of the fraction

2nd step: calculate the concentrations in mol/L of each component at equilibrium.

Molar Concentration Formula: straight C with straight m subscript equal to space numerator straight n degree sign space space mols over denominator volume space left parenthesis straight L right parenthesis end of fraction

PCl3 Cl2 PCl5
straight C with straight m subscript equal to numerator 0 comma 6 mol space over denominator 3 straight space L end of the fraction straight C with straight m subscript space end of subscript equal to 0 comma 2 mol space divided by straight L straight C with subscript straight m equal to numerator 0 comma 0120 mol space over denominator 3 straight space L end of fraction straight C with subscript m equal to 0 comma 004 mol space divided by straight L straight C with subscript straight m equal to numerator 0 comma 120 mol space over denominator 3 straight space L end of fraction straight C with subscript m equal to 0 comma 04 mol space divided by straight L

3rd step: replace the concentrations in the constant expression and calculate the value of Kç.

straight K with straight c subscript space equal to space numerator left square bracket PCl with 5 subscript right square bracket on denominator left square bracket PCl with 3 subscript right bracket right space x space bracket left square bracket Cl with 2 subscript right square bracket end of fraction equal to numerator 0 comma 04 mol space divided by straight L over denominator 0 comma 2 mol space divided by straight L straight space x space 0 comma 004 mol space divided by straight L space end of fraction straight K with straight c subscript space equal to space numerator 0 comma 04 mol space divided by straight L over denominator 0 comma 0008 mol squared space divided by straight L squared space end of fraction straight K with straight c subscript space equal to space 50 space left parenthesis mol divided by straight L right parenthesis to the minus 1 end power of exponential

Applications for equilibrium equilibrium

9. (Enem/2016) After their complete wear, the tires can be burned to generate energy. Among the gases generated in the complete combustion of vulcanized rubber, some are pollutants and cause acid rain. To prevent them from escaping into the atmosphere, these gases can be bubbled into an aqueous solution containing a suitable substance. Consider the substance information listed in the table.

equilibrium constants and n and n

Among the substances listed in the table, the one capable of most efficiently removing polluting gases is (a)

a) Phenol.
b) Pyridine.
c) Methylamine.
d) Potassium hydrogen phosphate.
e) Potassium hydrogen sulfate.

Correct answer: d) Potassium hydrogen phosphate.

the CO2, sulfur oxides (SO2 and SO3) and nitrogen oxides (NO and NO2) are the main polluting gases.

When they react with water present in the atmosphere, there is a acid formation that cause an increase in the acidity of the rain, which is why it is called acid rain.

The equilibrium constants given in the table are calculated by the ratio between the concentrations of products and reagents as follows:

straight K with straight c subscript space equal to space numerator left straight parenthesis Products parenthesis right square on denominator left square bracket Reagents right square bracket end of fraction

Note that the equilibrium constant is proportional to the concentration of products: the greater the quantity of products, the greater the value of Kç.

Note the first and last composite values ​​in the table for Kç:

pyridine 1 comma 3 straight space x 10 space to the minus 10 power of the exponential 0 comma 00000000013
Potassium hydrogen sulfate 3 comma 1 space x space 10 to the power of minus 2 end of exponential 0 comma 031

Comparing the two numbers, we see that the smaller the negative power, the greater the value of the constant.

To remove pollutants more efficiently, OH- to react with H ions+ present in acids through a neutralization reaction.

Among the substances presented, those that produce the hydroxyls necessary to neutralize acidic compounds are: pyridine, methylamine and potassium hydrogen phosphate.

To find out which compound is the most efficient, we observe the equilibrium constants: the higher the value of the constant, the greater the concentration of OH-.

Thus, the aqueous solution containing a substance suitable for this purpose is potassium hydrogen phosphate, as it is more basic and neutralizes acids more efficiently.

To learn more, read these texts.:

  • ionic balance
  • Neutralization reaction

10. (Enem/2009) Soaps are salts of long-chain carboxylic acids used in order to facilitate, during washing processes, the removal of substances of low water solubility, eg oils and fats. The following figure represents the structure of a soap molecule.

carboxylic acid salt

In solution, soap anions can hydrolyze water and thereby form the corresponding carboxylic acid. For example, for sodium stearate, the following balance is established:

hydrolysis

Since the carboxylic acid formed is poorly soluble in water and less efficient in removing fats, the pH of the medium must be controlled in such a way as to prevent the balance above being shifted to the right.

Based on the information in the text, it is correct to conclude that soaps work in a way:

a) More efficient at basic pH.
b) More efficient at acidic pH.
c) More efficient at neutral pH.
d) Efficient over any pH range.
e) More efficient at acidic or neutral pH.

Answer: a) More efficient at basic pH.

In the balance shown, we see that sodium stearate when reacting with water forms a carboxylic acid and hydroxyl.

The purpose of controlling the pH is not to allow the formation of carboxylic acid, and this is done by shifting the balance by changing the OH concentration-.

the more OH- in solution, there is a disturbance on the products side and the chemical system reacts by consuming the substance that had its concentration increased, in this case the hydroxyl.

Consequently, there will be the transformation of products into reagents.

Therefore, soaps work most efficiently at basic pH as excess hydroxyl shifts the balance to the left.

If the pH were acidic, there would be a higher concentration of H+ that would affect the balance by consuming OH- and the balance would act by producing more hydroxyl, shifting the balance to the left and producing more carboxylic acid, which is not of interest in the presented process.

Chemical equilibrium shift

11. (Enem/2011) Soft drinks have increasingly become the target of public health policies. Glue products contain phosphoric acid, a substance that is harmful to the fixation of calcium, the mineral that is the main component of the teeth matrix. Caries is a dynamic process of imbalance in the process of dental demineralization, loss of minerals due to acidity. It is known that the main component of tooth enamel is a salt called hydroxyapatite. The soda, due to the presence of sucrose, decreases the pH of the biofilm (bacterial plaque), causing the demineralization of the dental enamel. Salivary defense mechanisms take 20 to 30 minutes to normalize the pH level, remineralizing the tooth. The following chemical equation represents this process:

demineralization of tooth enamelGROISMAN, S. Impact of soda on teeth is assessed without taking it off the diet. Available in: http://www.isaude.net. Accessed on: 1 May 2010 (adapted).

Considering that a person consumes soft drinks daily, a process of dental demineralization may occur, due to the increased concentration of

a) OH, which reacts with Ca ions2+, shifting balance to the right.
b) H+, which reacts with the OH hydroxyls, shifting balance to the right.
c) OH, which reacts with Ca ions2+, shifting balance to the left.
d) H+, which reacts with the OH hydroxyls, shifting balance to the left.
e) Ca2+, which reacts with the OH hydroxyls, shifting balance to the left.

Correct answer: b) H+, which reacts with the OH hydroxyls, shifting balance to the right.

When the pH decreases it is because the acidity has increased, that is, the concentration of H ions+, as the statement says, there is the presence of phosphoric acid.

These ions react with OH- causing this substance to be consumed and, consequently, to shift the balance to the right, as the system acts by producing more of these removed ions.

The equilibrium shift between reactants and products occurred due to the decrease in the OH concentration-.

If the Ca ions2+ and oh- had the concentration increased, it would shift the balance to the left, as the system would react by consuming them and forming more hydroxyapatite.

12. (Enem/2010) Sometimes, when opening a soda, it is noticed that a part of the product quickly leaks from the end of the container. The explanation for this fact is related to the disturbance of the chemical balance existing between some of the product's ingredients according to the equation:
CO with 2 left parenthesis straight g right parenthesis subscript end of subscript space plus straight space H with 2 subscript straight O with left parenthesis straight l right parenthesis subscript end of subscript space right arrow over left arrow space H with 2 subscript CO with 3 left parenthesis aq right parenthesis subscript end of subscribed

Changing the previous balance, related to refrigerant leakage under the conditions described, results in:

a) CO release2 for the environment.
b) Raising the temperature of the container.
c) Raising the internal pressure of the vessel.
d) Elevation of CO concentration2 in the liquid.
e) Formation of a significant amount of H2O.

Correct answer: a) CO release2 for the environment.

Inside the bottle, the carbon dioxide was dissolved in the liquid due to high pressure.

When the bottle is opened, the pressure inside the container (which was greater) equals the pressure in the environment and, with that, there is an escape of carbon dioxide.

The equilibrium shift between reactants and products occurred due to the decrease in pressure: when the pressure decreases, the equilibrium shifts to the largest volume (number of moles).

The reaction shifted to the left and the CO2 which was dissolved in the liquid was released, leaking out when the bottle was opened.

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